# Thread: ddifferentiation using l'Hopitals rule

1. ## differentiation using l'Hopitals rule

Hello, I am trying to diffferentiate using l'Hopital's rule:
(e^(5+h)-e^5) / (h)
to find the limit of h tending towards zero.

I differentiate the numerator and denominator to get:

(e^(5+h) - e^5) / (1)

but I think I should be getting (e^(5+h)) / (1)
Can someone please tell me how the second e^5 is done away with ?

Thanks kindly.

2. ## Re: ddifferentiation using l'Hopitals rule

You don't differentiate using L'Hospital's Rule, you differentiate the top and the bottom because that's what you need to do to USE L'Hospital's Rule. Anyway

\displaystyle \displaystyle \begin{align*} \lim_{h \to 0}\frac{e^{5 + h} - e^5}{h} &= \lim_{h \to 0}\frac{e^5e^h - e^5}{h} \\ &= \lim_{h \to 0}\frac{e^5\left(e^h - 1\right)}{h} \\ &= e^5\lim_{h \to 0}\frac{e^h - 1}{h} \textrm{ which goes to }\frac{0}{0}\textrm{ so L'Hospital's Rule can be used} \\ &= e^5\lim_{h \to 0}\frac{\frac{d}{dx}\left(e^h - 1\right)}{\frac{d}{dx}\left(h\right)} \textrm{ by L'Hospital's Rule} \\ &= e^5\lim_{h \to 0}\frac{e^h}{1} \\ &= e^5\lim_{h \to 0}e^h \\ &= e^5 \cdot 1 \\ &= e^5 \end{align*}

3. ## Re: differentiation using l'Hopitals rule

Originally Posted by fran1942
Hello, I am trying to diffferentiate using l'Hopital's rule:
(e^(5+h)-e^5) / (h)
to find the limit of h tending towards zero.

I differentiate the numerator and denominator to get:

(e^(5+h) - e^5) / (1)

but I think I should be getting (e^(5+h)) / (1)
Can someone please tell me how the second e^5 is done away with ?

Thanks kindly.

Circular reasoning

4. ## Re: ddifferentiation using l'Hopitals rule

e^5 is a constant and differentiates to zero